Ribosomes.ribosomes are small structures where proteins are made.Although they are not enclosed within a membrane,they are frequently considered organelles. Hope it helps :)
Answer:
The answer to your question is 33.4 ml
Explanation:
Data
volume 1 = V1 = 42 ml
temperature 1 = T1 = 20°C
temperature 2 = T2 = -60°C
Volume 2 = V2 = x
Process
1.- Convert celsius to kelvin
T1 = 20 + 273 = 293°K
T2 = -60 + 273 = 233°K
2.- Use the Charles' law to solve this problem
Solve for V2
V2 = 
3.- Substitution
V2 = 
4.- Simplification
V2 = 
5.- Result
V2 = 33.4ml
The balanced chemical reaction is:
N2 + 3H2 = 2NH3
We are given the amount of hydrogen gas to be used in the reaction. This will be the starting point of the calculations.
24.0 mol H2 (2 mol NH3 / 3 mol H2 ) = 16 mol NH3
Therefore, ammonia produced from the reaction given is 16 moles.
The isotope that is more abundant, given the data is isotope Li7
<h3>Assumption</h3>
- Let Li6 be isotope A
- Let Li7 be isotope B
<h3>How to determine whiche isotope is more abundant</h3>
- Molar mass of isotope A (Li6) = 6.02 u
- Molar mass of isotope B (Li7) = 7.02 u
- Atomic mass of lithium = 6.94 u
- Abundance of A = A%
- Abundance of B = (100 - A)%
Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]
6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]
6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]
6.94 = [6.02A% + 702 - 7.02A%] / 100
Cross multiply
6.02A% + 702 - 7.02A% = 6.94 × 100
6.02A% + 702 - 7.02A% = 694
Collect like terms
6.02A% - 7.02A% = 694 - 702
-A% = -8
A% = 8%
Thus,
Abundance of B = (100 - A)%
Abundance of B = (100 - 8)%
Abundance of B = 92%
SUMMARY
- Abundance of A (Li6) = 8%
- Abundance of B (Li7) = 92%
From the above, isotope Li7 is more abundant.
Learn more about isotope:
brainly.com/question/24311846
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